In this answer, I'll give an explicit formula for the interpolating polynomial and their limit. As a result, I'll conclude that
(1) $p_k(x)-x$ is increasing as a function of $x$ on $(-\infty, 0]$, with $p_k(0)=1$, so $p_k(x)$ has a unique negative root $x_k$ and
(2) $p_k(x)-x$ is decreasing as a function of $k$ for $x$ fixed in $(-\infty, 0)$, so $x_k$ is increasing. Thus, $\lim_{k \to \infty} x_k$ exists, and Joachim Konig's computations show that the limit cannot be $-\gamma$.
First, though, we make some changes of variables. Put $q_n(x) = p_n(x)-x$, so $q_n(\tfrac{k(k-1)}{2}) = k$ for $1 \leq k \leq n+1$, you want to solve $q_n(x)=0$.
Put $r_n(z) = 2 q_n(\tfrac{z^2-1}{8})-1$. So $r_n(2k-1) = 2k-1$ for $1 \leq k \leq n+1$ and we want to solve $r_n(z)=-1$ for $z$ imaginary. Since $r_n(z)$ is a polynomial in $z^2$, it is an even function, so we have $r_n(z) = |z|$ for $z \in \{ -(2n+1), -(2n-1), \ldots, -3,-1,1,3, \ldots, 2n-1, 2n+1 \}$. The things we want to show about $r_n$ are
(1) $r_n(iy)$ is decreasing as a function of $y \in [0,\infty)$ and
(2) $r_n(iy)$ is decreasing as a function of $n$ for $y \in [0,\infty)$.
With those preliminaries out of the way, here is our formula for $r_n$:
$$r_n(z) = 1- \sum_{k=1}^n \frac{1}{2k-1} \prod_{j=1}^k \frac{(2j-1)^2 - z^2}{(2j)^2}. \qquad (\ast)$$
Thus,
$$r_n(iy) = 1- \sum_{k=1}^n \frac{1}{2k-1} \prod_{j=1}^k \frac{(2j-1)^2 + y^2}{(2j)^2}.$$
From this formula, properties (1) and (2) are clear.
I was going to write up a proof of $(\ast)$, but it got pretty long. The conceptual point is that $r_n(z-2) - 2 r_n(z) + r_n(z+2)$ is a polynomial of degree $2n-2$ vanishing at $\{ \pm 3, \pm 5, \ldots, \pm (2n-1) \}$, so it must be $c \prod_{j=2}^{n-1} ((2j-1)^2 - z^2)$ for some $c$, and we can evaluate $c$ by plugging in $z=1$. Then I have to unwind to get $r_n(z)$ itself, which is a pain.
Fedor Petrov, in comments, suggests a better approach to proving $(\ast)$. It is enough to show that $r_n(2m+1)=2m+1$ for $0 \leq m \leq n$. All the terms which differ between $r_n(2m+1)$ and $r_m(2m+1)$ are $0$, so it is enough to check that $r_{2m+1}(2m+1)=2m+1$. Fedor simplifies $r_{2m+1}(2m+1)$ to
$$\sum_{k=0}^m \frac{(-1)^{k-1}}{2k-1} \frac{(m+k)!}{k! k! (m-k)!} = \sum_{k=0}^m \frac{(-1)^{k-1}}{2k-1} \binom{2k}{k} \binom{m+k}{2k}.$$
We can drop the upper bound of the sum, since $\binom{m+k}{2k}=0$ for $k>m$.
We use Wilf's "snake oil method":
$$\sum_{m=0}^{\infty} x^m \sum_k \frac{(-1)^{k-1}}{2k-1} \binom{2k}{k} \binom{m+k}{2k} = \sum_k \frac{(-1)^{k-1}}{2k-1} \binom{2k}{k} \sum_{m=0}^{\infty} \binom{m+k}{2k} x^m$$
$$=\sum_k \frac{(-1)^{k-1}}{2k-1} \binom{2k}{k} \frac{x^k}{(1-x)^{2k+1}}. \quad (\dagger)$$
We have
$$\sum_{k=0}^{\infty} \frac{(-1)^{k-1}}{2k-1} \binom{2k}{k} y^k = \sqrt{1+4y}$$
so
$$(\dagger) = \frac{1}{1-x} \sqrt{1+\tfrac{4x}{(1-x)^2}} = \frac{1}{1-x} \frac{1+x}{1-x} = \frac{1+x}{(1-x)^2}.$$
The Taylor series of $\tfrac{1+x}{(1-x)^2}$ is $1+3x+5x^2+\cdots$ so, extracting the coefficient of $x^m$, we prove the result.
Finally, we note that
$$\frac{(2j-1)^2 -z^2}{(2j)^2} = 1 - 1/j + O(1/j^2)$$
for bounded $z$, so
$$\frac{1}{2k-1} \prod_{j=1}^k \frac{(2j-1)^2 - z^2}{(2j)^2} = O(1/k^2).$$
Thus, if we extend the sum $(\ast)$ to an infinite sum, it will converge uniformly on compact sets, and we get an entire function
$$\phi(z):=1- \sum_{k=1}^{\infty} \frac{1}{2k-1} \prod_{j=1}^k \frac{(2j-1)^2 - z^2}{(2j)^2}.$$
I've done a little experimenting. If you believe in Mathematica's ability to evaluate infinite sums, then $\phi(0) = 2/\pi$. Numerically, $\phi(z)$ seems to be very close to
$$\sqrt{z^2 + \tfrac{4}{\pi^2} \cos^2 \tfrac{\pi z}{2}},$$
but I don't think they are actually equal.
